Question

Match the following by using suitable identities.

• A. 103.02
• B. 3860
• C. 9801
• D. 9975

Answer 1

Case 1:
$(99{)}^2$ can be written as
$(100-1{)}^2$
Using the identity $(a-b{)}^2=a^2+b^2-2ab$,
$(100-1{)}^2=(100{)}^2+(1{)}^2-2\times 100\times 1$
$\Rightarrow (99{)}^2=10000+1-200=9801$

Case 2:
$(95\times 105)$ can be written as,
$(100-5)\times (100+5)$
Using the identity $(a+b)(a-b)=a^2-b^2$,
$(100+5)(100-5)=(100{)}^2-(5{)}^2$
$\Rightarrow 95\times 105=10000-25=9975.$

Case 3:
$10.1\times 10.2$ can be written as $(10+0.1)\times (10+0.2)$
Using the identity $(x+a)(x+b)=x^2+(a+b)x+ab$,
Here, x = 10, a = 0.1 and b = 0.2
$(10+0.1)(10+0.2)=(10{)}^2+(0.1+0.2)10+0.1\times 0.2$
$\Rightarrow 10.1\times 10.2=100+3+0.02=\mathrm{103.02.}$

Case 4:
$(69.3{)}^2-(30.7{)}^2$
Using the identity $a^2-b^2=(a+b)(a-b)$,
Here, a = 69.3 and b = 30.7
$(69.3{)}^2-(30.7{)}^2=(69.3+30.7)(69.3-30.7)$
$\Rightarrow (69.3{)}^2-(30.7{)}^2=100\times 38.6=3860.$